Question

当前状态是此代码：

while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['timeStamp'] . "</td>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['what'] . "</td>";
echo "<td>" . $row['salad'] . "</td>";
echo "<td>" . $row['tomatoes'] . "</td>";
echo "<td>" . $row['onions'] . "</td>";
echo "<td>" . $row['carrots'] . "</td>";
echo "<td>" . $row['hot'] . "</td>";
echo "<td>" . $row['cheese'] . "</td>";
echo "<td>" . $row['sauce'] . "</td>";
echo "</tr>";

沙拉，西红柿，洋葱，胡萝卜，热和奶酪这些行被保存为真和假，因为用户将它们作为复选框插入。我基本上想要的是在写入sql结果的表中显示复选框时为true或false

这有可能吗？让我知道您是否需要更多代码（例如整个php）。

Answer 1

如果我理解的没错，您可以简单地添加一个复选框并设置被检查的属性（如果值是true，如下所示）

echo '<td><input type="checkbox" value="true" ';
echo ($row['salad'])? 'checked' : '';
echo '></td>';

请注意，如果您不希望用户对其进行编辑或取消选中它，则可以添加已禁用属性。

Answer 2

假设您已为每个项目将值保存为db中的true或false

while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['timeStamp'] . "</td>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['what'] . "</td>";
echo "<td><input type='checkbox' " . ($row['salad']=='true' ? 'checked' :'') . "/></td>";
echo "<td><input type='checkbox' " . ($row['tomatoes']=='true' ? 'checked' :'') . "/></td>";
echo "<td><input type='checkbox' " . ($row['onions']=='true' ? 'checked' :'') . "/></td>";
echo "<td><input type='checkbox' " . ($row['carrots']=='true' ? 'checked' :'') . "/></td>";
echo "<td><input type='checkbox' " . ($row['hot']=='true' ? 'checked' :'') . "/></td>";
echo "<td><input type='checkbox' " . ($row['cheese']=='true' ? 'checked' :'') . "/></td>";
echo "<td><input type='checkbox' " . ($row['sauce']=='true' ? 'checked' :'') . "/></td>";
echo "</tr>";

在SQL中显示对与错选择查询为复选框

2 个答案: